Chapter 16 Aqueous Equilibria The intravenous (IV) solution in this bag might save a life. The...

Chapter 16 Aqueous Equilibria

The intravenous (IV) solution in this bag might save a life. The concentration of each solute in the solution is carefully selected to keep the total solute concentration within an optimal range and to maintain the pH of the blood. Because even slight deviations in blood pH can be fatal, the first treatment administered to an injured person is usually an intravenous solution.

Single acid (base) Salts and mixed salts (anions and cations). Central topic: How pH values can be affected and controlled by ions.

Assignment

16.5,16.14,16.26,16.31,16.38,

16.47,16.52,16.61,16.71,16.76.

Acidity/Basicity of Various Solutions

NaCl CuSO4 KBr

Characteristic colors of 10 mM water solutions of K2FeO4, KMnO4 and Fe2 (SO4)3. 1 = Permanganate; 2 = Potassium ferrate; 3 = Ferric chloride. Why?

How to modify the pH value of a solution?

Indicators: tell you how acidic or basic your

samples are by looking at it.

Salts in Water

• Ions as acids and bases

• The pH of a salt solution

• The pH of mixed solutions

Figure 16.1 Solutions of salts in water give rise to acidic, neutral, and basic solutions, as shown here by the color of the indicator bromothymol blue (see Table 16.3). From left to right are solutions of ammonium chloride, sodium chloride, and sodium acetate. Throughout this chapter, we use the Brønsted definitions of acids and bases.

+3

+4 2 3

+ - + -2

+ +2 2

- +3

NH (aq)+H O(1) NH + (aq)

Na (aq)+Cl (aq)+2H O(1) Na (aq)+Cl (aq)+

H

+ (aq)

Na (aq)+H O(1)+HCO (aq) Na (aq)+

OH (a

HCOOH(aq)+ (a

O

H

q)

) Oq

OH

Salts that contain the conjugate acids of weak bases produce acidic aqueous solutions.

All anions that are conjugate bases of weak acids act as proton acceptors,giving basic solutions.

Figure 16.2 A solution of titanium(III) sulfate is so acidic that it can releaseH2S from some sulfides.

2-4 n 2

2-2 6 4 n

3+ - + n+ n- 2-2 6-k k n 4

3+ - + 2-42

3+

3+

(aq)+SO (aq)+XS (aq)+H O(l)

(aq) {H O} +SO (aq)+XS (aq)

Ti (aq) {[H O] +OH ]}+kH +X +S +SO (aq)

Ti (aq)+ +OH +X +SO (

Ti

T

H S ( )g q)

i

a

Small, highly charged metal ionshave strong polarizing effect onwater so that they act as a proton donator, an acid.

Figure 16.3 In water, Al3 cations exist as hydrated

ions that can act as Brønsted acids. Although, for clarity, only four water molecules are shown here, metal cations typically have six H2O molecules attached to them.

Small, highly charged metal ions exert the greatest pull on the electrons.They look like a proton donator, an acid.

H+

H+

H+

2 6

n+ -

n

+-k k

+

2 6

(aq) {H O}

M (aq) {[H O] +OH ]} k

M

+ H

(A cation cannot be a base)

However,

• the cations of Group 1 and 2 metals and those of charge +1 from other groups are such weak Lewis acids that they DO NOT act as acids when hydrated. These metal ions are too large or have too low a charge to have an appreciable polarizing effect on water molecules that surround them. As a result, the hydrating water molecules do not readily release their protons.

Questions

• Is Na+ an acid in water?

• Is Ca2+ an acid in water?

No and No

Figure 16.4 The relative strengths of conjugate acids and bases have a reciprocal relation. (a) When a species has a high tendency to donate a proton, the resulting conjugate base has a low tendency to accept one. (b) When a species has a low tendency to donate a proton, the resulting conjugate base has a strong tendency to accept one. In the insets, a solid blue color represents the water molecules.

Strong Acid (Base) Very Weak Conjugate Base (Acid)

– most weak acids weak conjugate bases

– most weak bases weak conjugate acids

Ka=0.1 Kb=10-2 Kb=10-13 Ka=10-12

Ka=10-4 ~ 10-10 Kb=10-10 ~ 10-4

Kb=10-4 ~ 10-10 Ka=10-10 ~ 10-4

Only very few anions are acids

All anions that are conjugate bases of weak acids act as proton acceptors,giving basic solutions.

(Very weak bases Kb 0 neutral)

Exercise

• Decide whether aqueous solutions of (a) Ba(NO2)2; (b) CrCl3; (c) NH4NO3 are acidic, basic or neutral.

(a) Basic, for the NO2- is the conjugate base of a weak acid.

(b) Acidic, for Cr3+ is small and highly charged so that it can polarize water and release proton(s) from water.(c) Acidic, for the NH4

+ is the conjugate acid of a weak base (and NO3

- is the conjugate base of a strong acidneutral)

• (a) Ba(NO2)2; (b) CrCl3; (c) NH4NO3

Classroom Exercise

• Decide whether aqueous solutions of (a) Na2CO3; (b) AlCl3; (c) KNO3 are acidic, basic or neutral.

(a) Basic for the CO32- is the conjugate base of a weak acid.

(b) Acidic for Al3+ is small and highly charged so that it can polarize water and release proton(s) from water.(c) Neutral for the K+ is large and charge +1 so that it cannot polarize water.

• (a) Na2CO3; (b) AlCl3; (c) KNO3

Figure 16.5 The initial (left) and equilibrium (right) composition of a solution of a salt composed of the cation HB and the anion A, where HB is a weak acid and A is neutral. (left) The hypothetical initial situation that we imagine before deprotonation. (right) At equilibrium, the acidic cation is partially deprotonated, and the solution is acidic.

How to calculate the pH of an acidic solution

The pH of a Salt Solution(1) The initial molarity of the acidic ion is the molarity of the initially

completely protonated ion. We assume that the initial molarities of its conjugate base and H3O+ are 0.

(2) Write the increase in molarity of H3O+ as x mol/L and use the reaction stoichiometry to write the corresponding changes for the acidic ion and its conjugate base. Ignore H3O+ from the autoprotolysis of water; this approximation is valid if [H3O+] is substantially (about 10 times) greater than 10-7. Check the validity of this approximation at the end of the calculation.

(3) Write the equilibrium molarities of the species of x. (4) Express the acidity constant for the ion in terms of x and solve the

equations for x; assume that x is small, but check the validity of this approximation at the end of the calculation. If Ka is not available, obtain it from the value of Kb for the conjugate base by using Ka=Kw/Kb. Because x mol/L is the H3O+ molarity, the pH of the solution is –logx.

Follow the same procedure as above, except that now proton transfer from water to the ion results in the formation of OH- and the conjugate acid of the ion. We therefore use Kb ad the equilibrium table leads to a value of pOH.

An amphiprotic anion can act as both an acid and a base: pH = (pKa+pKb)/2

Acidic

Basic

Example • Estimate the pH of 0.15 M NH4Cl (aq). NH3: Kb = 1.8×10-5 (Table 15.3)

++ + 3 3

4 2 3 3 a4

[H O ][NH ]NH (aq)+H O(l) H O (aq)+NH (aq) K

[NH ]

4 3 3 NH H O NH

Initial 0.15 0 0

Change -

Equilibrium 0.15 -

x x x

x

x x

14w

5b

2 2

K 101.0 10K 1.8 10

+10 3 3

0.15- 0.154

10 6

63

=5.6 10

[H O ][NH ]5.6 10 = =

[NH ]

0.15 5.6 10 9.2 10

(less then 0.5% of 0.15, the approximation is valid)

[H O ]= 9.2 10 >> autoprotlysis mola

a

x xx

K

x

x

7

6

rity (1.0 10 ),

meaning autoprotolysis contribution can be neglected.

pH log[9.2 10 ] 5.04

Example • Estimate the pH of 0.10 M methylammonium chloride, CH3NH3Cl (aq).

CH3NH3+: Kb=3.6×10-4 (Table 15.3) or Ka = 2.8×10-11 (Table 16.1)

++ + 3 3 2

3 3 2 3 3 2 a +3 3

[H O ][CH NH ]CH NH (aq)+H O(l) H O (aq)+CH NH (aq) K

[CH NH ]

+3 3 3 3 2 CH NH H O CH NH

Initial 0.10 0 0

Change -

Equilibrium 0.10 -

x x x

x x

x

14w

4b

2 2

K 111.0 10K 3.6 10

+11 3 3 2

0.10- 0.10+3 3

11 6

63

=2.8 10

[H O ][CH NH ]2.8 10 = =

[CH NH ]

0.10 2.8 10 1.67 10


[H O ]= 1.67 10 >> autopro

a

x xx

K

x

x

7

6

tlysis molarity (1.0 10 ),


pH log[1.67 10 ] 5.78

Classroom Example • Estimate the pH of 0.10 M NH4NO3 (aq). NH3: Kb = 1.8×10-5 (Table 15.3)

++ + 3 3

4 2 3 3 a4

[H O ][NH ]NH (aq)+H O(l) H O (aq)+NH (aq) K

[NH ]

4 3 3 NH H O NH

Initial 0.10 0 0

Change -

Equilibrium 0.10 -

x x x

x

x x

14w

5b

2 2

K 101.0 10K 1.8 10

+10 3 3

0.10- 0.104

10 6

63

=5.6 10

[H O ][NH ]5.6 10 = =

[NH ]

0.10 5.6 10 7.5 10


[H O ]= 7.5 10 >> autoprotlysis mola

a

x xx

K

x

x

7

6

rity (1.0 10 ),


pH log[7.5 10 ] 5.12

Example • Estimate the pH of 0.15 M calcium acetate, Ca(CH3CO2)2 (aq). CH3CO2

-: Ka = 1.8×10-11 (Table 15.3)

- -3 2 3 CH CO OH CH COOH

Initial 0 0

Change -

Equilibrium 0.30 -

0.30

x x x

x x

x

14w

5a

2 2

K 101.0 10b K 1.8 10

10 30.30- 0.30

3 2

10 5

- 5

=5.6 10

[CH COOH][OH ]5.6 10 = =

[CH CO ]

0.30 5.6 10 1.3 10


[OH ]= 1.3 10 >> autoprotlys

x xx

K

x

x

7

5

is molarity (1.0 10 ),


pOH log[1.3 10 ] 4.89 pH=9.11

32 3 2 3 b

3 2

[CH COOH][OH ]H O(l) CH CO (aq) CH COOH(aq) OH (aq) K

[CH CO ]

Example • Estimate the pH of 0.1 M potassium heptate, KC6H5CO2 (aq). C6H5COOH:

Ka = 6.5×10-5

- -6 5 2 6 5 C H CO OH C H COOH

Initial 0.10 0 0

Change -

Equilibrium 0.10 -

x x x

x x

x

14w

5a

2 2

K 101.0 10b K 6.5 10

10 6 50.10- 0.10

6 5 2

10 6

- 6

=1.54 10

[C H COOH][OH ]1.54 10 = =

[C H CO ]

0.1 1.54 10 7.3 10


[OH ]= 7.3 10 >> autopr

x xx

K

x

x

7

6

otlysis molarity (1.0 10 ),


pOH log[7.3 10 ] 5.14 pH=8.86

6 52 6 5 2 6 5 b

6 5 2

[C H COOH][OH ]H O(l) C H CO (aq) C H COOH(aq) OH (aq) K

[C H CO ]

Classroom Example • Estimate the pH of 0.02 M potassium fluoride, KF (aq). HF: Ka =

3.5×10-4

- - F OH FH

Initial 0.02 0 0

Change -

Equilibrium 0.02 -

x x x

x

x x

14w

4a

2 2

K 111.0 10b K 3.5 10

110.02- 0.02

11 7

- 7

=2.86 10

[FH][OH ]2.86 10 = =

[F ]

0.02 2.86 10 7.56 10


[OH ]= 7.56 10 >> autoprotlysis m

x xx

K

x

x

7

7

olarity (1.0 10 ),


pOH log[7.56 10 ] 6.12 pH=7.88

2 b

[FH][OH ]H O(l) F (aq) FH(aq) OH (aq) K

[F ]

Mixed Solution: Example • Estimate the pH of a solution that is 0.5 M HNO2 and 0.1 M KNO2 (aq).

HNO2 :Ka = 4.3×10-4 (Table 15.3)

2 3 2 HNO H O NO

Initial 0.5 0

Change -

Equilibrium 0.

0.

5 -

1

x x x

x x

0.1+x

(0.1 )4 3 2 0.10.5- 0.50

2

3

+ 3 73

[H O ][NO ]4.3 10 = =

[HNO ]

2.2 10


[H O ]= 2.2 10 >> autoprotlysis molarity (1.0 10 ),

meaning autoprotolysis co

x x xx

x

x

3

ntribution can be neglected.

pH log[2.2 10 ] 2.66

3 22 2 3 2 a

2

[H O ][NO ]HNO (aq)+H O(l) H O (aq)+NO (aq) K

[HNO ]

Mixed Solution: Example • Estimate the pH of a solution that is 0.3 M CH3NH2 and 0.146 M

CH3NH3Cl(aq). CH3NH2 :Kb = 3.6×10-4 (Table 15.3)

- +(0.146 )4 3 3 0.146

b 0.3- 0.303 2

4

- 4 7

[OH ][CH NH ]3.6 10 = K =

[CH NH ]

7.4 10


[OH ]= 7.4 10 >> autoprotlysis molarity (1.0 10 ),

meaning autop

x x xx

x

x

4

rotolysis contribution can be neglected.

pOH log[7.4 10 ] 3.13 pH=10.87

- +- + 3 3

3 2 2 3 3 b3 2

[OH ][CH NH ]CH NH (aq)+H O(l) OH (aq)+CH NH (aq) K

[CH NH ]

+3 2 3 3 CH NH OH CH NH

Initial 0.30 0

Change -

Equilibrium 0.30 -

0.1

46

x x x

x x

0.146+x

Classroom Example • Estimate the pH of a solution that is 0.01 M HClO (aq) and 0.2 mM

NaClO(aq). HClO :Ka = 3.0×10-8 (Table 15.3)

-4

3 HClO H O ClO

Initial 0.01 0

Change -

Equilibrium 0

2.0 1

.01-

0

x x x

x

-4 2.0 1 + 0 xx

-4 -4(2.0 10 )8 3 2.0 10a 0.01- 0.01

6

+ 6 73

[H O ][ClO ]3.0 10 = K =

[HClO]

1.5 10



meaning aut

x x xx

x

x

6

oprotolysis contribution can be neglected.

pH log[1.5 10 ] 5.82

32 3 a

[H O ][ClO ]HClO(aq)+H O(l) H O (aq)+ClO (aq) K

[HClO]

Titrations

• Strong acid-strong base titrations

• Weak acid-strong base and strong acid-weak base titrations

• Titrating a polyprotic acid

• Indicators

Titration: A titrant adds H3O+ (or OH-) to and analyte so thata neutral solution (stoichiometric point) is obtained (the initial pH of the analyte is thus found).

Figure 16.6 The variation of pH during the titration of a strong base (the analyte) with a strong acid (the titrant). This curve is for 25.00 mL of 0.250 M NaOH(aq) titrated with 0.340 M HCl(aq). The stoichiometric point occurs at pH 7 (point S ). The other points are discussed in the text.

slow

fast

Figure 16.7 The variation of pH during a typical titration of a strong acid (the analyte) with a strong base (the titrant). The stoichiometric point occurs at pH 7.

slow

fast

Figure 16.8 The composition of the solutions initially (left) and at the stoichiometric point (right) in the titration of a strong acid (the analyte) with a strong base (the titrant).

Step 1: calculate the moles of H3O+ (if the analyte is a strong acid) or OH- (if the analyte is a strong base) in the original analyte solution from its molarity andvolume.Step 2: calculate the moles of H3O+ (if the titrant is a strong acid) or OH- (if the titrant is a strong base) in the volume of titrant added.

Step 3: write the chemical equation for the neutralization reaction and use reaction stoichiometry to find the moles of H3O+ (or OH- if the analyte is strong base) that remain in the analyte

solution. Each mole of H3O+ ions reacts with 1 mol OH- ions; therefore subtrsct the number of of moles of H3O+ or OH- ions that have reacted from the initial number ofmoles of H3O+ or OH- ions.

Step 4: Determine the concentration of hydronium or hydroxide.

Step 5: Find the pH (pOH).

Calculating points on the pH curve for a strong acid-strong base titration (before stoichiometric point)

Titrant: 0.34 M HCl(aq); analyte: 25 mL, 0.25 M NaOH. After addition of 5.0 mL of titrant, pH=?

1. The initial pOH of the analyte: -log(0.25)=0.602pH=13.40. The amount of OH- is 0.025 L×0.25 M = 6.25 mmol.

2. The amount of H3O+ supplied by the titrant is: 0.005 L×0.34 M = 1.70 mmol

3. After the reaction of all the hydronium ions added, the amount of hydroxide ion remaining is 6.25 – 1.70 = 4.55 mmol.

4. Because the total volume of the solution is now 30 mL, the molarity of hydroxide ion is 4.55 mmol/30 mL = 0.152 M

5. pOH=-log(0.152)=0.82 pH = 13.18.

Example: titration (before stoichiometric point)






5. pOH=-log(0.0814)=1.09 pH = 12.91.

Classroom Exercise






5. pOH=-log(0.0586)=1.232 pH = 12.77.

Stoichiometric Point

Titrant: 0.34 M HCl(aq); analyte: 25 mL, 0.25 M NaOH. How much titrant is needed to reach stoichiometric point?


2. The amount of H3O+ supplied by the titrant is assumed to be x at the stoichiometric point, then

6.25 mmol = x×340 mM x = 18.38 mL

For strong acids-strong base titration, the pH is 7.0, but for othercases, the pH may be not 7.0 at the stoichiometric point!

Calculating points on the pH curve for a strong acid-strong base titration (after stoichiometric point)




3. After the reaction of all the hydroxide ions, the amount of hydronium ion remaining is 6.60 – 6.25 = 0.35 mmol.

4. Because the total volume of the solution is now 44.4 mL, the molarity of hydronium ion is 0.35 mmol/44.4 mL = 7.9 mM

5. pH=-log(0.0079)=2.10.

Classroom Exercise (pH after stoichiometric point)

Titrant: 0.34 M HCl(aq); analyte: 25 mL, 0.25 M NaOH. After addition of 25 mL of titrant, pH=?



3. After the reaction of all the hydroxide ions, the amount of hydronium ion remaining is 8.5 – 6.25 = 2.25 mmol.

4. Because the total volume of the solution is now 50 mL, the molarity of hydronium ion is 2.25 mmol/50 mL = 45 mM

5. pH=-log(0.045)=1.35.

NaOH

HCl

0.25 MNaOH

0.34 MHCl

pH

5 ml

0 ml 25 ml 13.40

25 ml 13.18

10 ml 25 ml 12.91

12 ml 25 ml 12.77

18.38 ml 25 ml 7.00

19.4 ml 25 ml 2.10

25 ml 25 ml 1.35

Titration of NaOH by HCl

Animation of Acid-Base Titration

http://science.csustan.edu/chem/titrate/Acid/titrateZ.swf

Follow the instructions and try a few times with different initial concentrations/volumes of acid and base. Click the yellow button on the valve to see the generation of the titration curve.(The titration curve might grow rather slow. Be patient.)



Weak acid-strong base and strong acid-weak base titrations

Examples:

Chemical EngineeringGeneral Industry or Lab Use (e.g. cleaning agents and surface treatments with liquid cleaners, degreasers, strippers, passivators, etchants, solutions and additives for cleaning and surface preparation. )

Biochemistry (intracellular/extracellular equilibrium, neuron activation)Medicine (IV, pharmacology, toxicology)

Printed Circuit Board (PCB)Integrated CircuitsSemiconductor ManufactureNanomaterials (nanoparticles)Photography

HCOOH+NaOHNaHCO2+H2O

Na+ + HCO2-

A strong base dominates a weak acid; the solution at the stoichiometric point is basic.

Weak acid-strong base titrations

HCO2- + H2O HCOOH +OH-

HClO2+NH4OH NH4Cl+H2O

Cl- + NH4+

A strong acid dominates a weak base; the solution at the stoichiometric point is acidic.

Strong acid-weak base titrations

NH4+ + H2O NH3 +H3O+

Figure 16.9 The composition of the solutions in the course of the titration of a weak acid with a strong base. From left to right: at the start of the titration, just before the stoichiometric point, at the stoichiometric point, and well after the stoichiometric point. At the stoichiometric point, the OH ions are mainly those arising from proton transfer from H2O to A.

(basic)

Calculating the pH during a titration of a weak acid or a weak base

Step 1: calculate the moles of weak acid (or weak base) in the original analyte solution from its molarity and volume.

Step 3: write the chemical equation for the neutralization reaction and use reaction stoichiometry to find:

Step 4: find the molarity of the conjugate acid and base in solution by dividing the moles of each species by the total volume of the solution.

Weak acid-strong base titration: the moles of conjugate base formed by the reaction of the acid with the added base, and the moles of weak acid remaining. Weak base-strong acid titration: the moles of conjugate acid formed by the reaction of the base with the added acid, and the moles of weak base remaining.

Step 2: calculate the moles of H3O+ (if the analyte is a weak acid) or OH- (if the analyte is a weak base) in the in the volume of titrant added.

Step 5: for a weak acid, use the equilibrium table to find the H3O+ concentration and in each case, assume that the contribution of the autoptolysis of water to the pH is insignificant if the pH is less than 6 or larger than 8. Step 6: find pH from the hydronium concentration. For a weak base, use an equilibrium table to find the OH- concentration and find pOHpH.

HCOOH+NaOHNaHCO2+H2O

Na+ + HCO2-

HCO2- + H2O HCOOH +OH-

1 2 3

4

5

pOHpH

6

Kb

Calculating the pH during a titration of a weak acid

weak base

(at stoichiometric point)

(No HCOOH left)

Basic!

HA + MOH MA + H2O

M+ + A-

A- + H2O HA +OH-

1 2 3

4

5

pOHpH

6

Kb


weak base


(No HA left)

Basic!

MA MB MA-MA

-

MA- 0 0

-x

x x

x x

MA- - x2

w

aA

Kxb M x KK

Example: Weak Acid-Strong Base Titration Estimate the pH at the stoichiometric point of the titration

of 25.00 mL of 0.1 M HCOOH (aq) with 0.15 M NaOH (aq).

2 2HCO (aq) H O(l) HCOOH(aq) OH (aq)

2 2HCOOH(aq)+NaOH(aq) NaHCO (aq)+H O(l)

Moles of HCOOH: 0.0025 L × 0.1 mol/L = 2.5 mmol.

Because all HCOOH has reacted to form HCO2-, the moles of HCO2

-

at the stoichiometric point is also 2.5 mmol.

From the chemical equation, 1 mol OH- is needed for 1 mol HCOOH:

The moles of NaOH required is then: 2.5 mmol/0.15 M = 16.7 mL.

The total volume of the solution at the stoichiometric point is 25 mL + 16.7 mL = 41.7 mL. The molarity of sodium formate is 2.5 mmol/41.7 mL = 0.06 M.Because the formate ion is a weak base, the equilibrium to consider is:

2 2HCO (aq) H O(l) HCOOH(aq) OH (aq) Because the formate ion is a weak base, the equilibrium to consider is:

From Table 15.3, Ka=1.8 ×10-4 for formic acidKb =5.6 ×10-11.

- -2 HCO HCOOH OH

Initial 0.06 0 0

Change -

Equilibrium 0.06 -

x x x

x x

x

2 2110.06- 0.06

2

11 6

- 6 7

[HCOOH][OH ]5.6 10 = =

[HCO ]

0.06 5.6 10 1.8 10



meaning autoproto

x xx

x

x

6

lysis contribution can be neglected.

pOH log[1.8 10 ] 5.7 8p .24 H= 6 Basic!

Figure 16.10 The pH curve for the titration of a weak acid with a strong base. This curve is for the titration of 25.00 mL of 0.100 M HCOOH(aq) with 0.150 M NaOH(aq). The stoichiometric point (S ) occurs on the basic side of pH 7 because the anion HCO2

is a base.

Classroom Exercise: Weak Acid-Strong Base Titration

Estimate the pH at the stoichiometric point of the titration of 25.00 mL of 0.01 M HClO (aq) with 0.02 M KOH (aq).

2ClO (aq) H O(l) HClO (aq) OH (aq)

2HClO(aq)+KOH(aq) KClO (aq)+H O(l)

Moles of HClO: 0.0025 L × 0.01 mol/L = 0.25 mmol.

Because all HClO has reacted to form ClO-, the moles of ClO-


From the chemical equation, 1 mol OH- is needed for 1 mol HClO:

The moles of KOH required is then: 0.25 mmol/0.02 M = 12.5 mL.

The total volume of the solution at the stoichiometric point is 25 mL + 12.5 mL = 37.5 mL. The molarity of KClO is 0.25 mmol/37.5 mL = 6.67 mM.Because the ClO- is a weak base, the equilibrium to consider is:

From Table 15.3, Ka=3.0×10-8 for HClOKb =3.33 ×10-7.- -

-3

ClO HClO OH

Initial 6.67 10 0 0

Change -

Equilibri

x x x

-3um 6.67 10 - x x x

2 2

3 3

7

6.67 10 - 6.67 10

10 -5

- -5 7

[HClO][OH ]3.33 10 = =

[ClO ]

6.67 3.33 10 4.71 10



meani

x xx

x

x

-5

ng autoprotolysis contribution can be neglected.

pOH log[4.71 10 ] 4.33 9.6H= 7p Basic!

2ClO (aq) H O(l) HClO (aq) OH (aq) Because the ClO- is a weak base, the equilibrium to consider is:

HA + MOH MA + H2O

M+ + A-

HA+ H2O A- + H3O+

1 2 3

4

5

pH = -log(x)

6

Ka


weak acid

(before stoichiometric point)

MA MB MA-MA

-

MA-MA-MB 0

-xMA- + x x

x x

MA-MB-x( )

A

A B

x M x

a M M xK

Acidic!

Example: Weak Acid-Strong Base Titration(before stoichiometric point)

Estimate the pH: 5.0 mL of 0.15 M NaOH (aq) is added to 25.00 mL of 0.1 M HCOOH (aq).

- -2 2HCOOH(aq)+ OH (aq) HCO (aq)+H O(l)


The amount of OH- in 5.0 mL of the titrant is 5.0 mL × 0.15 M = 0.75 mmol.

From the chemical equation,

0.75 mmol of OH- produces 0.75 mmol of HCO2- and leaves:

2.5 mmol – 0.75 mmol = 1.75 mmol of HCOOH.

The total volume of the solution at this stage is 25 mL + 5 mL = 30 mL, so the molarities of acid and conjugate base are: HCOOH: 1.75 mmol/30 mL = 0.0583 M. HCO2

-: 0.75 mmol/30 mL = 0.025 M

+2 2 3HCOOH(aq) H O(l) HCO (aq)+H O (aq)

The proton transfer equilibrium in water is:

From Table 15.3, Ka=1.8 ×10-4 for formic acid.

+ -3 2 HCOOH H O HCO

Initial 0.0583 0

Change -

Equil

0.

ibrium

025

x x x 0.0583 - 0.025+x x x

+(0.025+x)4 2 3 0.0250.0583- 0.0583

4

+ 4 73

[HCO ][H O ]1.8 10 = =

[HCOOH]

4.2 10



meaning autopr

x xx

x

x

4

otolysis contribution can be neglected.

pH log[4.2 10 ] 3.38 Acidic, of cause(before stoichiometric point)

Example: Weak Acid-Strong Base Titration(before stoichiometric point)

Estimate the pH: 10.0 mL of 0.15 M NaOH (aq) is added to 25.00 mL of 0.1 M HCOOH (aq).








-: 1.5 mmol/35 mL = 0.0428 M





Initial 0.0286 0

Change -

0.0

428

x x

Equilibrium 0.0286 - 0.0428+

x

x x x

+(0.0428+x)4 2 3 0.04280.0286- 0.0286

4

+ 4 73

[HCO ][H O ]1.8 10 = =

[HCOOH]

1.2 10



meaning auto

x xx

x

x

4

protolysis contribution can be neglected.

pH log[1.2 10 ] 3.92 Less acidic, of cause(more NaOH has been added)

Classroom Exercise: Weak Acid-Strong Base Titration

(before stoichiometric point)) Estimate the pH: 15.0 mL of 0.15 M NaOH (aq) is added to

25.00 mL of 0.1 M HCOOH (aq).








-: 2.25 mmol/40 mL = 0.0563 M





Initial 0.00625 0

Change -

0.0563

Equilibrium

x x x 0.00625 - 0.0563+x x x

+(0.0563+x)4 2 3 0.05630.00625- 0.00625

5

+ 5 73

[HCO ][H O ]1.8 10 = =

[HCOOH]

2.0 10



meaning a

x xx

x

x

5

utoprotolysis contribution can be neglected.

pH log[2.0 10 ] 4.7 Acidic, of cause(before stoichiometric point)

Figure 16.12 The pKa of an acid can be determined by carrying out a titration of the weak acid with a strong base (or vice versa) and locating the pH of the solution after the addition of half the volume of titrant needed to reach the stoichiometric point.

3 2a 3

a

[H O ][HCO ]K =[H O ]

[HCOOH]

pH = pK

HA + MOH MA + H2O

M+ + A-

A- + H2O HA +OH-

1 2 3

4

5

pOHpH

6Kbweak base

(after stoichiometric point)

(No HA left)

Basic!

MA MB MA-MA

-

MA- 0 MB - MA-

-x x x

MA- - x( )B wA

aA

x x M M Kb M x KK

x x+MB - MA-


HPr = Propionic Acid

Weak-acid-strong-base

titration curve

The Four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak

Acid-Strong Base Titration Curve

1. The initial pH is higher.

2. A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point.

3. The pH at the equivalence point is greater than 7.00.

4. The steep rise interval is less pronounced.

HClO2+NH4OH NH4Cl+H2O

Cl- + NH4+

A strong acid dominates a weak base; the solution at the stoichiometric point is acidic.

Strong acid-weak base titrations

NH4+ + H2O NH3 +H3O+

B + HA BHA

BH+ + A-

BH+ + H2O B +H3O+

1 2 3

4

5

pH6Ka

weak acidAcidic!


Calculating the pH during a titration of a weak base

Estimate the pH at the stoichiometric point of the titration of 25.00 mL of 0.02 M NH3 (aq) with 0.015 M HCl (aq). NH4

+: Ka =5.6×10-10.

+ +4 2 3 3NH (aq) H O(l) NH (aq) H O (aq)

3 2 4 2NH (aq)+HCl(aq)+H O(l) NH Cl(aq)+H O(l)

Moles of NH3: 0.025 L × 0.02 mol/L = 0.5 mmol.

Because all NH3 has reacted to form NH4+, the moles of NH4

+


From the chemical equation, 1 mol H+ is needed for 1 mol NH3:

The moles of HCl required is then: 0.5 mmol/0.015 M = 33.33 mL.

The total volume of the solution at the stoichiometric point is 25 mL + 33.33 mL = 58.33 mL. The molarity of NH4Cl is 0.5 mmol/58.33 mL = 8.57 mM.

Because the NH4+ is a weak acid, the equilibrium to consider is:

Example: Strong Acid-Weak Base Titration

Ka=5.6 ×10-10 for NH4+.

+ +4 3 3

-3

NH NH H O

Initial 8.57 10 0 0

Change - x x

-3

Equilibrium 8.57 10 -

x

x x x

2 2

-3

+10 3 3 1000

8.57+ 8.57 10 -4

13 6

+ 6 73

[NH ][H O ]5.6 10 = =

[NH ]

8.57 5.6 10 2.19 10



meani

x xx

x

x

6

ng autoprotolysis contribution can be neglected.

pH log[2.19 10 ] 5.66 Acidic!

+ +4 2 3 3NH (aq) H O(l) NH (aq) H O (aq)

Because the NH4+ is a weak acid, the equilibrium to consider is:

Figure 16.11 A typical pH curve for the titration of a weak base with a strong acid. The stoichiometric point (S) occurs on the acidic side of pH 7 because the salt formed by the neutralization reaction has an acidic cation.

The four Major Differences Between a Weak Acid-Strong Base Titration Curve and a Weak

Base-Strong Acid Titration Curve

1. The initial pH is above 7.00.

2. A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point.

3. The pH at the equivalence point is less than 7.00.

4. Thereafter, the pH decreases slowly as excess strong acid is added.

Titrating a Polyprotic Acid

A-

H+

H+

H+

A-

H+

H+

A-

H+

A-

More than one stoichiometric point can be detected.

Titration of H3PO4 (aq)

3 4 2 4 2H PO (aq)+OH (aq) H PO (aq)+H O(1)

22 4 4 2H PO (aq)+OH (aq) HPO (aq)+H O(1)

2 34 4 2HPO (aq)+OH (aq) PO (aq)+H O(1)

1 mol

1 mol

1 mol

1 mol

1 mol

1 mol

Figure 16.13 The variation of the pH of the analyte solution during the titration of a triprotic acid (phosphoric acid) and the major species present in solution. SP1, SP2, and SP3 indicate the stoichiometric points. Points A, C and E are the points at which pH pKa for each deprotonation. Points B, D, and F are discussed in the text.

Example What volume of 0.02 M NaOH (aq) is required to reach the (a) first; (b)

second stoichiometric point in a titration of 30 mL of 0.01 M H3PO4 (aq)?

1 mol

1 mol

1 mol

1 mol

0.3 mmol y mL ×0.02 M =0.3 mmol

y= 15 mL

0.3 mmolz mL ×0.02 M =0.3 mmol z= 15 mL x+y+z= 45 mL

1 mol 1 mol

30 mL ×0.01 M =0.3 mmol

x mL ×0.02 M =0.03 mmol

x= 15 mL

3 4 2 4 2H PO (aq)+OH (aq) H PO (aq)+H O(1)

22 4 4 2H PO (aq)+OH (aq) HPO (aq)+H O(1)

2 34 4 2HPO (aq)+OH (aq) PO (aq)+H O(1)

x+y= 30 mL

Classroom Exercise What volume of 0.01 M NaOH (aq) is required to reach the (a) first; (b)

second stoichiometric point in a titration of 25 mL of 0.01 M H2SO3 (aq)?

2 3 3 2H SO (aq)+OH (aq) HSO (aq)+H O(1)

23 3 2HSO (aq)+OH (aq) SO (aq)+H O(1)

1 mol

1 mol

1 mol

1 mol

25 mL ×0.01 M =0.025 mmol

x mL ×0.01 M =0.025 mmol x= 25 mL

0.025 mmol y mL ×0.01 M =0.025 mmol y= 25 mL x+y= 50 mL

Figure 16.14 A commercially available automatic titrator. The stoichiometric point of the titration is detected by a sudden change in pH; the pH is monitored electrically, using a technique described in Investigating Chemistry 18.1. The pH can be plotted as the reaction proceeds, as shown on the left.

Acid-Base Indicators• An acid-base indictor is a weak acid that has one

color in its acid form (HIn) and another color in its conjugate base form (In-). (The light absorption characteristics of HIn are different from those of In-.)

InH

In-

++ 3

2 3 In

[H O ][In ]HIn(aq)+H O(1) H O (aq)+In (aq) =

[HIn]K

The end point of a titration is defined as the point at which the concentrations of the acid and base forms of the indicator are equal: [HIn] = [In-]. At this point, the color of the indicator ishalfway between the colors of its acid and base forms.

+In 3 In=[H O ] pH=pKK

Figure 16.15 The stoichiometric point of an acid-base titration may be detected by the color change of an indicator. Here we see the colors of solutions containing a few drops of phenolphthalein (酚酞 ) at (from left to right) pHs of 7.0, 8.5, 9.4 (its end point), 9.8, and 12.0. At the end point, equal amounts of the conjugate acid and base forms of the indicator are present.

End point

Figure 16.16 The same dye is responsible for the red of poppies (a) and the blue of cornflowers（矢車菊） (b). The color difference is a consequence of the more acidic sap of poppies.

Natural Indicators

Figure 16.17 Ideally, an indicator should have a sharp color change close to the stoichiometric point of the titration, which is at pH 7 for a strong acid-strong base titration. However, the change in pH is so abrupt that phenolphthalein can also be used. The color change of methyl orange, however, would give a less accurate result.

Figure 16.18 Phenolphthalein can be used to detect the stoichiometric point of a weak acid-strong base titration, but methyl orange would give a very inaccurate indication of the stoichiometric point. The pH curves are superimposed on approximations to the colors of the indicators in the neighborhoods of their end points.

Figure 16.19 Methyl orange can be used for a weak base-strong acid titration. Phenolphthalein would be inappropriate because its color change occurs well away from the stoichiometric point. The pH curves are superimposed on approximations to the colors of the indicators in the neighborhoods of their end points.

A Question

Q: How many methods can you list to evaluate the pH of a solution?

A: At least three: (1) pH meter, (2) automatic indicator, (3) acid-base indictors.

Other possible methods: Absorption spectroscopy, NMR, electrochemical method etc.

Buffer Solutions

• The action of buffers

• Selecting a buffer

• Buffer capacity

Solutions that resist changes in pH when small amounts of strong acid or base are added.

The Action of Buffers

An acid buffer solution consists of a weak acid and its conjugate base: HA/A-; it has pH<7.0.

An base buffer solution consists of a weak base and its

conjugate acid: BH+/B; it has pH>7.0.

Blood and other cell fluids buffered at pH = 7.4.Oceans are buffered at pH=8.4.

Acid Buffer Action

+3 2 3 3 2CH COOH(aq)+H O(1) H O (aq)+CH CO (aq)

Acid buffer action: the weak acid transfers protons to the OH- ions from addedstrong base. The conjugate base of the weak acid accepts protons from the H3O+ ions supplied by a strong acid.

HA H3O+ CH3COOH pH change very small.

BOH OH- consumes H3O+CH3COOH CH3CO2-

pH change very small.

Base Buffer Action

Base buffer action: the conjugate acid of the weak base transfers protons to the OH- ions from added strong base. The weak base accepts protons from the H3O+ ions supplied by a strong acid.

HA H3O+ promotes NH3 to protonate and consumed by OH-


BOH OH- stimulates NH3 to deprotonate NH4+


+3 2 4NH (aq)+H O(1) NH (aq)+OH (aq)

Figure 16.20 Buffer action depends on the donation of protons by the weak acid molecules, HA, when a strong base is added and the acceptance of protons by the conjugate base ions, A, when a strong acid is added. In the inset, a solid blue color represents water.

Design a Buffer with Specified pH

+3 a

+3 a

a a

initiala

initial

[HA][H O ]=K

[A ]

[HA]log[H O ]= logK log

[A ]

[HA] [A ]pH=pK log =pK +log

[A ] [HA]

[base]pH pK +log .

[acid]Henderson-Hasselbalch equation

This is a problem of mixed solutions (both an acid and the salt of itsconjugate base are contained).

+ -+ 3

2 3 a

[H O ][A ]HA(aq)+H O(1) H O (aq)+A (aq) K =

[HA]

Acid buffer:

Example Calculate the pH of a buffer solution that is 0.04 M NaCH3CO2 (aq)

and 0.08 M CH3COOH (aq) at 25 oC. CH3COOH: pKa = 4.75 (Table 15.3)

+ 33 a

3 2

+ 33 a

3 2

3 3 2a a

3 2 3

[CH COOH][H O ]=K

[CH CO ]

[CH COOH]log[H O ]= logK log

[CH CO ]

[CH COOH] [CH CO ]pH=pK log =pK +log

[CH CO ] [CH COOH]

0.04pH 4.75+log 4.45.

0.08

++ 3 3 2

3 2 3 3 2 a3

[H O ][CH CO ]CH COOH(aq)+H O(1) H O (aq)+CH CO (aq) K =

[CH COOH]

Design a Buffer with Specified pH

-b +

-b +

+

b b+

initialb

initial

[B][OH ]=K

[BH ]

[B]log[OH ]= logK log

[BH ]

[B] [BH ]pOH=pK log =pK +log

[BH ] [B]

[acid]pOH pK +log pH=14-pOH.

[base]

Base buffer:- +

- +2 b

[OH ][BH ]B(aq)+H O(1) OH (aq)+BH (aq) K =

[B]

Example Calculate the pH of a buffer solution that is 0.04 M NH4Cl (aq) and 0.03

M NH3 (aq) at 25 oC. NH3: pKb = 4.75 (Table 15.3)

- 3b

4

- 3b

4

3 4b b

4 3

[NH ][OH ]=K

[NH ]

[NH ]log[OH ]= logK log

[NH ]

[NH ] [NH ]pOH=pK log =pK +log

[NH ] [NH ]

0.04pOH 4.75+log 4.87 pH=9.13.

0.03

+ -+ - 4

3 2 4 b3

[NH ][OH ]NH (aq)+H O(1) NH (aq)+OH (aq) K =

[NH ]

Classroom Exercise Calculate the pH of a buffer solution that is 0.15 M HNO2 (aq) and 0.2 M

NaNO2 (aq) at 25 oC. HNO2: pKa = 3.37 (Table 15.3)

+ 23 a

2

+ 23 a

2

2 2a a

2 2

[HNO ][H O ]=K

[NO ]

[HNO ]log[H O ]= logK log

[NO ]

[HNO ] [NO ]pH=pK log =pK +log

[NO ] [HNO ]

0.2pH 3.37+log 3.49.

0.15

++ 3 2

2 2 3 2 a2

[H O ][NO ]HNO (aq)+H O(1) H O (aq)+NO (aq) K =

[HNO ]

Calculating the pH change when acid or base added to a buffer

3 2 initiala

3 initial

[CH C5.4

O ] 0.10pH pK +log =4.75+log .

[CH COOH] .5

0 02

-3 3 2 2CH COOH(aq)+OH CH CO (aq)+H O(1)

1.2 g (0.03 mol) of NaOH is dissolved in 500 mL of a buffer solution that is 0.04 M NaCH3CO2 (aq) and 0.08 M CH3COOH (aq) with pH=4.45. Calculate the pH of the resulting solution and the change in pH. Assume volume is not changed.

3 Moles of CH COOH = 0.5 L 0.08 M = 0.04 mol

0.01 mol3 0.5 LMolarity of CH COOH = = 0.02 M

3-

1 mol CH COOH- -

1 mol OH The amount of acetic acid that reacts: 0.03 mol OH = 0.03 mol OH

-3 2Moles of CH CO 0.5 L 0.04 M + 0.03 mol = 0.05 mol

3 The amount of CH COOH remaining is 0.04 - 0.03 mol = 0.01 mol

-3 2 3 3 2CH COOH(aq)+H O(1) H O (aq)+CH CO (aq)

- 0.05 mol3 2 0.5 LMolarity of CH CO = 0.1 M

Figure 16.21 A buffer solution contains a sink for protons (a weak base) that are supplied when a strong acid is added and a source of protons (a weak acid) to supply to a strong base that is added. The joint action of the source and the sink keeps the pH constant when a small amount of either strong acid or strong base is added.

Figure 16.22 When conjugate acid and base are present at similar concentrations, the pH changes very little as more strong base (or strong acid) is added. As the inset shows, the pH lies between pKa 1 for a wide range of concentrations.

Solubility Equilibria

• The solubility product

• The common-ion effect

• Predicting precipitation

• Selective precipitation

• Dissolving precipitates

• Complex ions and solubilities

• Qualitative analysis

Figure 16.23 A pollution control officer collects samples of river water in England. The water will be tested for the presence of heavy metal ions by adding a solution containing anions such as sulfide ions. If a precipitate appears, further tests will be conducted to identify the particular ions present.

3+ 2 3+ 2 2 32 3 sp

16.11

Bi S (s) 2Bi (aq)+3S (aq) K =[Bi ] [S ]

Case Study 16 (a) Patients undergoing surgery often need to be supplied with several kinds of intravenous solutions. These doctors are operating on a patient’s liver, a procedure that requires the administration of three different intravenous solutions and a blood transfusion.

Case Study 16 (b) Three solutions commonly given intravenously: lactated Ringer’s solution, 0.9% sodium chloride, and 5% dextrose (glucose). The first two help to control electrolyte levels, the third maintains the blood sugar level, and all three help to maintain blood volume.

Common Ion Effect

+ +spAgCl(s) Ag (aq)+Cl (aq) K =[Ag ][Cl ]

Figure 16.24 If the concentration of one of the ions in solution is increased, then the concentration of the other is decreased to maintain a constant value of Ksp. (a) The cations (pink) and anions (green) in solution. (b) When more anions are added (together with their accompanying spectator ions, which are not shown), the concentration of cations decreases. In other words, the solubility of the original compound is reduced by the presence of a common ion. In the insets, a solid blue color represents the water molecules.

Figure 16.25 (a) A saturated solution of zinc acetate in water. (b) When acetate ions are added (as solid sodium acetate in the spatula shown in (a)), the solubility of the zinc acetate is significantly reduced, and additional zinc acetate precipitates.

Figure 16.26 The solution in the middle is at equilibrium. The solution on the left is supersaturated, so it forms a precipitate until solid and solute are in equilibrium. In the solution on the right, solid has been added to pure water. As the solid begins to dissolve, the solute concentration is lower than the equilibrium concentration. In this case, the solid dissolves until the equilibrium concentration is reached.

Predicting Precipitation

3 2 3 2

2+2

2+ 2+ 22 sp

2+ 2 2 3sp

Pb(NO ) (aq)+2KI(aq) 2KNO (aq)+PbI (s)

Pb (aq)+2I (aq) PbI (s)

PbI (s) Pb (aq)+2I (aq) K =[Pb ][I ]

Q =[Pb ][I ] =0.1×(0.1) =1×10

Figure 16.27 When a lead(II) nitrate solution is added to a solution of potassium iodide, yellow lead(II) iodide immediately precipitates.

Selective Precipitation

Dissolving Precipitates

3+3

2+ 23 3

2-3 2 3

2 3 2 2

Fe(OH) (s) Fe (aq)+3OH (aq)

ZnCO (s) Zn (aq)+CO (aq)

CO (aq)+ 2HC1(aq) H CO (aq)+2C1 (aq)

H CO (aq) H O(1)+CO (g)

2+ 2

23 2 3

CuS(s) Cu (aq)+S (aq)

3 S (aq)+ 8 HNO (aq) 3S(s)+2NO(g)+4H O(1)+6NO (aq)

Complex Ions and Solubilities

+ +3 3 2

+ +sp

++ + 3 2

3 3 2 f + 23

Ag (aq)+2NH (aq) Ag(NH ) (aq)

AgC1(s) Ag (aq)+C1 (aq) K =[Ag ][C1 ] (A)

[Ag(NH ) ]Ag (aq)+2NH (aq) Ag(NH ) (aq) K = (B)

[Ag ][NH ]

Qualitative Analysis

2+ 2-4 4

+ +3 3 2

+2 2 3 2 4

+3 2 3

Pb (aq)+CrO (aq) PbCrO (s)

Ag (aq)+2NH (aq) Ag(NH ) (aq)

Hg C1 (s)+2NH (aq) Hg(1)+HgNH C1(s)+NH (aq)+C1 (aq)

Ag(NH ) (aq)+C1 (aq)+2H O(aq)

AgC1(s)+2NH

+

4 2

+ 2-2 2 3

(aq)+2H O(1)

H S(aq)+H O(1) 2H O (aq)+S (aq)

Figure 16.28 The sequence for the analysis of cations by selective precipitation. (a) The original solution contains Pb2, Hg2

2, Ag, Cu2, and Zn2 ions. Addition of HCl(aq) precipitates AgCl, Hg2Cl2, and PbCl2, which can be removed by decanting and filtration. (b) Addition of H2S to the remaining solution precipitates CuS, which can also be removed. (c) Making the resulting solution basic by adding ammonia precipitates ZnS.

Figure 16.29 When an aqueous solution of ammonia is added to a silver chloride precipitate, the precipitate dissolves. However, when ammonia is added to a precipitate of mercury(I) chloride, mercury metal and mercury(II) ions are formed by disproportionation and the mass turns gray. Left to right: silver chloride in water, silver chloride in aqueous ammonia, mercury(I) chloride in water, mercury(I) chloride in aqueous ammonia. The different responses of AgCl and Hg2Cl2 to the addition of ammonia allow them to be distinguished in a mixture.

Assignment for Chapter 16

• 16.5,16.14,16.26,16.31,16.38,

• 16.47,16.52,16.61,16.71,16.76.

Chapter 16 Aqueous Equilibria The intravenous (IV) solution in this bag might save a life. The...

Documents

Transcript of Chapter 16 Aqueous Equilibria The intravenous (IV) solution in this bag might save a life. The...